Length of a geodesic between $x$ and $y$ on a Riemann manifold that is compact

Question

I'm new to Riemann Geometry and I read on Wikipedia in the section called "Riemann manifolds are metric spaces" that

Assuming the manifold is compact, any two points x and y can be connected with a geodesic whose length is d(x,y).

I understand how geodesic length is defined on Riemann manifold but I'm not entirely clear how compactness gives us the length of the geodesic. Could someone share some insights?

Answer 1

This is the theorem of Hopf Rinow.

https://en.wikipedia.org/wiki/Hopf%E2%80%93Rinow_theorem

If the manifold is not compact, it is not true: Consider $R^2-\{0\}$ with the standard flat metric, there does not exists a geodesic between $(x,0)$ and $(y,0)$ if $x<0, y>0$.

Answer 2

There is a relation between the length of curves on Riemannian manifolds and the distance function: the distance between any two points is defined as the $\inf$ over all lengths of rectifiable curves joining the two points.

On a compact manifold it can be shown that there always exists, in fact, a curve which realizes this distance (this is true on a larger class of manifolds, but on compact ones it's easier to see). The proof is a bit too elaborate for reproducing it here. It is then possible to show (using methods from the calculus of Variations) that such a curve has to satify a certain differential equation (the variational ODE for the lenght functional), which is the one obeyed by geodesics.

(Some prefer to define a geodesic as curve which locally realizes the distance between any two points on it. With that definition you have an additional step to relate this kind of curves to those satisfying the differential equation).